Average pay in the San Francisco metropolitan area was 19 percent above the national average in 2006, the highest among the 78 metropolitan areas studied by the National Compensation Survey (NCS), the Bureau of Labor Statistics of the U.S. Department of Labor reported.

In contrast, pay was lowest in the Brownsville, Texas metropolitan area with a pay relative of 78, meaning Brownsville workers earned an average of 78 cents for every dollar earned by workers nationwide. Using data from the NCS, pay relatives-a means of assessing pay differences-are available for each of the 9 major occupational groups within 78 metropolitan areas, as well as averaged across all occupations for each area.

A pay relative is a calculation of pay-wages, salaries, commissions, and production bonuses-for a given metropolitan area relative to the nation as a whole. The calculation controls for differences among areas in occupational composition, establishment and occupational characteristics, and the fact that data are collected for areas at different times during the year. Simple pay comparisons calculating the ratio of the average pay for an area versus the entire United States in percentage terms would not control for interarea differences in occupational composition and other factors, which may have a significant effect on pay relatives.

The pay relative in 2006 for workers in construction and extraction occupations in the San Francisco area was 122, meaning the pay in San Francisco for that occupational group averaged 22 percent more than the national average pay for that occupational group. By contrast, the pay relative for workers in construction and extraction occupations in the Brownsville, Texas area was 67, meaning pay for workers in those occupations averaged 33 percent less than the national average. Pay relatives calculated for all occupations were significantly different from the national average in 68 of the 78 areas.